Problem: In the calm waters of an inlet, Manon's boat moves with a velocity (speed and direction) vector $\vec{b_1} = (3,4)$. Once it enters the main river channel, however, Manon observes that it is now moving with a velocity vector $\vec{b_2} = (1,1)$. (Speeds are in meters per second, or $\text{m}/\text{s}$.) What is the speed of the river's current?
Consider vector $\vec c$ (depicted below), which represents the current. We can imagine how it would cause Manon's boat to slow down and change direction. It is reasonable to assume that the velocity of the current added with the initial velocity of Manon's boat equals the resultant velocity of Manon's boat. $\vec c + \vec{b_1} = \vec{b_2}$ We can now solve for $\vec c$. $\begin{aligned} \vec c + \vec{b_1} &= \vec{b_2}\\\\ \vec c &= \vec{b_2} - \vec{b_1}\\\\ \vec c &= (1\hat i + 1\hat j) - (3\hat i + 4\hat j)\\\\ \vec c &= -2\hat i + (-3)\hat j \end{aligned}$ We can find the magnitude of $\vec c$ (i.e., the speed of the current) using the Pythagorean theorem. $\begin{aligned} \| \vec c \|^2 &= (-2)^2 + (-3)^2\\\\ \| \vec c \| &= \sqrt{4 + 9}\\\\ \| \vec c \| &= \sqrt{13}\\\\ \| \vec c \| &\approx 3.6 \text{ m}/\text{s} \end{aligned}$ $\vec c$ is pointing into the third quadrant, so we can find its direction (call it $\theta~$ ) using the arctangent function and adding $180^\circ$. $\begin{aligned} \tan \theta &= \dfrac{-3}{-2}\\\\ \theta &= \arctan{\left ( \dfrac{3}{2} \right )} \\\\ \theta &= 56^\circ \end{aligned}$ Adding $180^\circ$ to this result gives us the actual direction, $236^\circ$ (rounded to the nearest degree). The speed of the current is $3.6 \text{ m}/\text{s}$. The direction of the current is $236^\circ$.